All categories

3 years ago

Sulfuric acid, H2SO4, can be neutralized by sodium hydroxide, NaOH. The unbalanced equation is:

Chemistry

1 answer:

kap26 [50]3 years ago

8 0

Answer:

Sulfuric acid, H2SO4, can be neutralized by sodium hydroxide, NaOH.

Explanation:

The balanced chemical equation of the reaction is:

$H_2SO_4(aq)+2NaOH(aq)->Na_2SO_4(aq)+2H_2O(l)$

A student who was asked to balance the reaction wrote the following:

H2SO4(aq) + Na2OH(aq) →Na2SO4(aq) + H3O(l)

This is wrong equation.

Because, the formula of sodium hydroxide is NaOH only and it is not $Na_2OH$ .

Because the valency of sodium will not exceed one and hydroxide ion has valency is also one.

H2SO4(aq) + Na2OH(aq) →Na2SO4(aq) + H3O(l)

for this equation, the left side mass of reactants is:

161.0g

Right side mass of products is:

161.0 g

But H3O(l) will not exist and $H_3O^+(aq)$ will exist.

You might be interested in

When an alkaline earth metal, a, reacts with a halogen, x, the formula of the covalent compund formed should be a2x?

Ierofanga [76]

Alkaline earth metal are the elements present in II group in the periodic table and are known as 'Metals' and have a charge of +2.

Alkaline earth metals - Be , Mg Ca, Sr , Ba, Ra

Halogens are present in VII A group in the periodic table and are 'Non-metals' and have a charge of -1.

Halogens - F, Cl, Br, I, At

When Alkaline earth metal (metals) combine with Halogens (non-metals) the compound formed will be ionic compound and the formula of the compound will be based on the charges of the element.

When we write the formula of the ionic compound the charges of the elements get criss crossed.

For example - Mg (Alkaline earth metal) have a charge of +2 and Cl (Halogen) have a charge of -1 and when they combine to form the formula their charges get criss crossed and we will get $Mg_{1}Cl_{2}$ or $MgCl_{2}$

When an alkaline earth metal, A, reacts with a halogen, X, the formula of the Ionic compound formed should be $AX_{2}$

3 0

3 years ago

A rigid tank contains 1.40 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to l

sergey [27]

Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

$PV=nRT$

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between pressure and number of moles of gas will be:

$\frac{P_1}{P_2}=\frac{n_1}{n_2}$

where,

$P_1$ = initial pressure of gas = 24.5 atm

$P_2$ = final pressure of gas = 5.30 atm

$n_1$ = initial number of moles of gas = 1.40 moles

$n_2$ = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

$\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}$

$n_2=0.301mol$

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

8 0

3 years ago

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c

uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

$I_{2} (g) + Cl_{2} (g) -----> 2 ICl (g)$

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

$K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}$

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = $\frac{(2x)^{2}}{(0.437-x) (0.269-x)}$

81.9 = $\frac{(2x)^{2}}{x^{2} -0.706x + 0.118}$

$(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]$

$4x^{2} = 81.9x^{2} -57.8x +9.66$

$77.9x^{2} -57.8x+9.66 = 0$

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

$[ICl]_{eq} = 2 ( 0.254) = 0.508 M$

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0

3 years ago

Explain how a plant is adapted for photosynthesis

snow_lady [41]

Answer:

the utencils of a plant is called photosynthesis and your wc

4 0

3 years ago

How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t

ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

= 2.75 x 10²³ molecules XeF₆.

8 0

3 years ago